Let us start by considering ordinary three dimensional Euclidean space. The distance $r$ between the origin and a point $P(x, y, z)$ is given by

$$r^2=x^2+y^2+z^2$$

This equation comes from the Pythagoras’ Theorem which is the cornerstone of Euclidean Geometry.  We may write the equation for the distance in matrix form as

$$r^2=\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Here the identity matrix is the metric tensor for Euclidean space. It is called the Euclidean metric and has signature +++. It tells us how to measure distances (and also angles etc.) in Euclidean space. Other spaces will have different metrics as we will see shortly. In two dimensions the last equation simplifies to 

$$r^2=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}G\begin{pmatrix}x\\y\end{pmatrix}$$

Where we represented the Euclidean metric in two dimensions by $G$. Now suppose in this two dimensional space we have a vector with components $x$ and $y$ and we rotate the axes so that the components become $x^{\prime}$ and $y^{\prime}$. Then the new components are related to the old ones via the matrix $R$ describing this rotation:

$$\begin{pmatrix}x^{\prime}\\y^{\prime}\end{pmatrix}=R\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a&b\\c&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

The fact that under rotations the distance $r$ does not change i.e. is invariant means the the metric tensor and rotation matrix are related by the following equation

$$G=R^TGR$$

We now substitute the actual matrices into this equation and perform the matrix multiplication to get

$$\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}$$

This gives us the following three equations

$$a^2+c^2=1$$

$$b^2+d^2=1$$

$$ab+cd=0$$

Since $\sin^2\theta+\cos^2\theta=1$ these equations are satisfied by setting

$$a=d=\cos\theta$$

and

$$\mathbin{-}b=c=\sin\theta$$

Thus in Euclidean space in two dimensions a rotation moves a point along a circle centered at the origin and the rotation matrix is 

$$R=\begin{pmatrix}\cos\theta&\mathbin{-}\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

Now let us move from ordinary three-dimensional Euclidean space to the four-dimensional world of Minkowski spacetime. The interval between an event $(ct, x, y, z)$ and the origin is given by

$$s^2=c^2t^2\mathbin{-}x^2\mathbin{-}y^2\mathbin{-}z^2$$

Notice we have three minus signs unlike the Euclidean case where the signs are all positive. We say the geometry of Minkowski spacetime is pseudo-Euclidean. We may write the equation for the interval in matrix form as

$$s^2=\begin{pmatrix}ct&x&y&z\end{pmatrix}\begin{pmatrix}ct\\\mathbin{-}x\\\mathbin{-}y\\\mathbin{-}z\end{pmatrix}=\begin{pmatrix}ct&x&y&z\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&\mathbin{-}1&0&0\\0&0&\mathbin{-}1&0\\0&0&0&\mathbin{-}1\end{pmatrix}\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}$$

The 4$\times$4 matrix is the metric for Minkowski space. It is called the Minkowski metric and has signature $+\mathbin{-}\mathbin{-}\mathbin{-}$. However we can also write the interval as 

$$s^2=\mathbin{-}c^2t^2+x^2+y^2+z^2$$

This means the Minkowski metric has an equivalent signature $\mathbin{-}+++$ and we need to choose one of the two conventions. In this course we will use $+\mathbin{-}\mathbin{-}\mathbin{-}$ exclusively. Four dimensions can be hard to work with so we suppress $y$ and $z$ and just work with time and one space dimension $(x)$. The simplified interval is

$$s^2=\begin{pmatrix}ct&x\end{pmatrix}\begin{pmatrix}1&0\\0&\mathbin{-}1\end{pmatrix}\begin{pmatrix}ct\\x\end{pmatrix}=\begin{pmatrix}ct&x\end{pmatrix}G^{\prime}\begin{pmatrix}ct\\x\end{pmatrix}$$

We represent the Minkowski metric in these two dimensions by $G^{\prime}$. Now let us perform a rotation in Minkowski space. Suppose we rotate the axes so the components $ct$ and $x$ becomes respectively $ct^{\prime}$ and $x^{\prime}$. Then calling our rotation matrix $L$ we have

$$\begin{pmatrix}ct^{\prime}\\x^{\prime}\end{pmatrix}=L\begin{pmatrix}ct\\x\end{pmatrix}=\begin{pmatrix}e&f\\g&h\end{pmatrix}\begin{pmatrix}ct\\x\end{pmatrix}$$

The fact that the interval is invariant under rotations means that 

$$G^{\prime}=L^TG^{\prime}L$$

thus we have

$$\begin{pmatrix}1&0\\0&\mathbin{-}1\end{pmatrix}=\begin{pmatrix}e&g\\f&h\end{pmatrix}\begin{pmatrix}1&0\\0&\mathbin{-}1\end{pmatrix}\begin{pmatrix}e&f\\g&h\end{pmatrix}=\begin{pmatrix}e^2\mathbin{-}g^2&ef\mathbin{-}hg\\ef\mathbin{-}hg&f^2\mathbin{-}h^2\end{pmatrix}$$

Again we get three equations this time they are

$$e^2\mathbin{-}g^2=1$$

$$f^2\mathbin{-}h^2=\mathbin{-}1$$

$$ef\mathbin{-}hg=0$$

Since $\cosh^2\psi\mathbin{-}\sinh^2\psi=1$ these three equations are satisfied by

$$e=h=\cosh\psi$$

$$f=g=\sinh\psi$$

Thus in two-dimensional Minkowski space a rotation (called a Lorentz boost) moves a point along a hyperbola centered at the origin and the rotation matrix is given by

$$L=\begin{pmatrix}\cosh\psi&\sinh\psi\\\sinh\psi&\cosh\psi\end{pmatrix}$$